One expression that is equivalent to this sum is 18+12

There are 45 questions on your math exam. You answered 8/10 of them correctly. How many questions did you answer correctly?

liraira [26]

Answer:

36 questions were correct.

Step-by-step explanation:

In order to find how many questions correct, you have to multiply it :

$\frac{8}{10} \times 45 = 36$

4 0

3 years ago

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Look at this diagram.

mixer [17]

9514 1404 393

Answer:

138°

Step-by-step explanation:

The angles of the linear pair are supplementary.

∠MLO = 180° -∠MLJ = 180° -42°

∠MLO = 138°

5 0

3 years ago

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2. Which of the following equations are perpendicular to 2y = -3x + 1 I. II. III. y=-x-1 - 2x + 3y = -5 2x + 3y = 2 (A) I only (

Novay_Z [31]

The equation in given as ;

2y = -3x + 1

This can be written as ;

y= -3/2 x + 1/2

This means the equation has a gradient of -3/2

Let this slope , be , ---------m1

For perperdicular lines , the product of their slopes = -1 .This means if the other line has a gradient of m2 then : m1 * m2 = -1

So from the answers :

i) y= 2/3 x - 1 the slope is 2/3

m2 = 2/3

m1 * m2 = -1 -------check the if this is true by using the two values of gradient as;

-3/2 * 2/3 = - 1 ------ This is true-----equation i

II.

-2x + 3y = -5

3y = 2x -5

y= 2/3 x -5/3 -----m2 here is 2/3

m1*m2 = -1

-3/2 * 2/3 = -1 -----this is true , so ----equation ii

iii)

2x + 3y = 2

3y = -2x + 2

y= -2/3 x + 2/3 -----m2 = -2/3

m1*m2 = -1

-3/2 * -2/3 = 1 -----this is not true,,,equation iii is not perpendicular to our equation.

so, equation i and ii are perpendicular to our equation .

Answer : B i and ii only

4 0

1 year ago

Why is my life such a math problem?

polet [3.4K]

Answer:

IDK

Step-by-step explanation:

7 0

3 years ago

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Find the volume v of the described solid s. the base of s is an elliptical region with boundary curve 4x2 + 9y2 = 36. cross-sect

Tasya [4]

$4x^2+9y^2=36\iff\dfrac{x^2}9+\dfrac{y^2}4=1$

defines an ellipse centered at $(0,0)$ with semi-major axis length 3 and semi-minor axis length 2. The semi-major axis lies on the $x$ -axis. So if cross sections are taken perpendicular to the $x$ -axis, any such triangular section will have a base that is determined by the vertical distance between the lower and upper halves of the ellipse. That is, any cross section taken at $x=x_0$ will have a base of length

$\dfrac{x^2}9+\dfrac{y^2}4=1\implies y=\pm\dfrac23\sqrt{9-x^2}$
$\implies \text{base}=\dfrac23\sqrt{9-{x_0}^2}-\left(-\dfrac23\sqrt{9-{x_0}^2}\right)=\dfrac43\sqrt{9-{x_0}^2}$

I've attached a graphic of what a sample section would look like.

Any such isosceles triangle will have a hypotenuse that occurs in a $\sqrt2:1$ ratio with either of the remaining legs. So if the hypotenuse is $\dfrac43\sqrt{9-{x_0}^2}$ , then either leg will have length $\dfrac4{3\sqrt2}\sqrt{9-{x_0}^2}$ .

Now the legs form a similar triangle with the height of the triangle, where the legs of the larger triangle section are the hypotenuses and the height is one of the legs. This means the height of the triangular section is $\dfrac4{3(\sqrt2)^2}\sqrt{9-{x_0}^2}=\dfrac23\sqrt{9-{x_0}^2}$ .

Finally, $x_0$ can be chosen from any value in $-3\le x_0\le3$ . We're now ready to set up the integral to find the volume of the solid. The volume is the sum of the infinitely many triangular sections' areas, which are

$\dfrac12\left(\dfrac43\sqrt{9-{x_0}^2}\right)\left(\dfrac23\sqrt{9-{x_0}^2}\right)=\dfrac49(9-{x_0}^2)$

and so the volume would be

$\displaystyle\int_{x=-3}^{x=3}\frac49(9-x^2)\,\mathrm dx$
$=\left(4x-\dfrac4{27}x^3\right)\bigg|_{x=-3}^{x=3}$
$=16$

6 0

3 years ago